A runner is jogging steadily at 7.7km/hr. When the runner is 3.6km from the finish line, a bird begins flying from the runner to the finish line at 23.1km/hr (3 times as fast as the runner). When the bird reaches the finish line it turns around and flies back to the runner. How far does the bird travel? Assume that the bird occupies only one point in space and it can turn without loss of speed.

b) After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the beginning (including the first encounter)?

(a) It takes the bird 3.6/23.1 = 0.1558 h to teach the finish line. The runner is then 2.4 km from the finish line. They meet again 1.8 km from the finish line, since the bird travels 3/4 of the 2.4 km and the man 1/4.

The total distance travelled by the bird is 3.6 + 1.8 = 5.4 km

(b) It takes the man 2.6/7.7 = 0.3377 h to reach the finish line. The bird travels three times as far in the same interval, back and forth. That would be 10.80 km